In three-dimensional space,the equation $3y + 4z = 0$ represents:

In the following case,find the distance of the given point from the corresponding given plane.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

$A$ variable plane passes through a fixed point $(3, 2, 1)$ and meets the $x, y,$ and $z$ axes at $A, B,$ and $C$ respectively. $A$ plane is drawn parallel to the $yz$-plane through $A$,a second plane is drawn parallel to the $zx$-plane through $B$,and a third plane is drawn parallel to the $xy$-plane through $C$. Then,the locus of the point of intersection of these three planes is:

The equation of the plane passing through $(4,4,0)$ and perpendicular to the planes $2x+y+2z+3=0$ and $3x+3y+2z-8=0$ is

For what value of $k$ do the planes $kx + 4y + z = 0$,$4x + ky + 2z = 0$,and $2x + 2y + z = 0$ intersect in a single line?

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is